what is the boiling point of a solution of .01 mole of glucose(C6H22o6) in 200 g of water. kb=0.512C/m.
To find the boiling point of the solution, we need to use the equation for boiling point elevation, which is given by:
∆Tb = kb * m
Where:
∆Tb = boiling point elevation
kb = molal boiling point constant (provided as 0.512 °C/m for water)
m = molality of the solution
To calculate the molality of the solution, we need to know the number of moles of solute (glucose, C6H12O6) and the mass of the solvent (water, H2O).
Given the following information:
- Number of moles of solute (glucose) = 0.01 mol
- Mass of solvent (water) = 200 g
- Molar mass of solute (glucose) = 180.16 g/mol
First, let's calculate the molality (m) of the solution:
Molality (m) = moles of solute / mass of solvent (in kg)
Mass of solvent (water) needs to be converted to kg:
Mass of solvent (water) = 200 g = 0.2 kg
Now, let's calculate the molality (m):
m = 0.01 mol / 0.2 kg
m = 0.05 mol/kg
Next, we can plug in the values into the boiling point elevation equation to find the ∆Tb:
∆Tb = kb * m
∆Tb = 0.512 °C/m * 0.05 mol/kg
∆Tb = 0.0256 °C
Finally, we can add the boiling point elevation (∆Tb) to the boiling point of pure water (100 °C) to find the boiling point of the solution:
Boiling point of solution = boiling point of pure water + ∆Tb
Boiling point of solution = 100 °C + 0.0256 °C
Boiling point of solution = 100.0256 °C
So, the boiling point of the solution of 0.01 mole of glucose (C6H12O6) in 200 g of water is approximately 100.0256 °C.